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Riemannian – Lecture 15

Dr. Emma Carberry

September 14, 2015

Sectional curvature

Recall that as a (1, 3)-tensor, the Riemann curvature endomorphism of the Levi-Civita connec- tion ∇ is

R(X,Y )Z = ∇X (∇Y Z) − ∇Y (∇X Z) − ∇[X,Y ]Z. It measures the failure of the to be locally isometric to Euclidean . Starting with the , there are various simplifications of this tensor one can define. An important one is , because it

• is the natural generalisation of Gauss curvature of the

• completely determines the Riemann curvature tensor

A two-dimensional subspace π of TpM is called a 2- to M at p.

Sectional curvature is defined on tangent 2-planes π ⊂ TpM. We will define the sectional curvature of a tangent 2-plane π in terms of a for π. In order to obtain an expression which is independent of this choice of basis, we will need a normalisation factor.

Definition 15.1. For linearly independent vectors X,Y ∈ TpM, let

Q(X,Y ) = ( of the with sides X and Y )2 = hX,Xi hY,Y i − hX,Y i2 .

Definition 15.2. Given any p ∈ M and any tangent 2-plane π to M at p, the sectional curvature of π is the hR(X,Y )Y,Xi Rm(X,Y,Y,X) sec(π) := = , Q(X,Y ) Q(X,Y ) where {X,Y } denotes a basis of π. In components Kij := sec (span{∂i, ∂j})

1 Independence of basis choice

Proposition 15.3. Sectional curvature sec(π) is well-defined, independent of the choice of basis {X,Y } for π.

Proof: If {X,˜ Y˜ } is another basis for π, then

X˜ = aX + bY, Y˜ = cX + dY, with non-zero ad − bc.

An easy computation gives

Q(X,˜ Y˜ ) = (ad − bc)2Q(X,Y ).

Using of Rm,

Rm(X,˜ Y,˜ Y,˜ X˜) = Rm(aX + bY, cX + dY, cX + dY, aX + bY ) = (ad − bc) Rm(X, Y, cX + dY, aX + bY ) = (ad − bc)2Rm(X,Y,Y,X) so sec(π) is independent of the choice of basis {X,Y }.

We have seen that the sectional curvature of M is a real-valued defined on the set of all tangent 2-planes of M.

Although sectional curvature seems simpler than the curvature tensor Rm, it encapsulates the same information.

Lemma 15.4. Given sec(π) for all tangent 2-planes π ⊂ TpM, we can algebraically determine the curvature tensor Rm at p.

Proof: The idea is to use the symmetry of the Riemann curvature tensor. Set X,Y,Z,W ∈

TpM, and consider the polynomial in t defined by

f(t) = Rm(X + tW, Y + tZ, Y + tZ, X + tW ) − t2(Rm(X,Z,Z,X) + Rm(W, Y, Y, W ))

Due to the in each term, we can write f in terms of sectional (and the function Q which is given by the inner product).

The coefficient of t2 in f(t) is

Rm(X,Y,Z,W ) + Rm(X,Z,Y,W ) + Rm(W, Z, Y, X) + Rm(W, Y, Z, X).

2 Recall that Rm is skew-symmetric in each of the first and last 2 entries   Rm(X,Y,Z,W ) = −Rm(Y,X,Z,W ),  Rm(X,Y,Z,W ) = −Rm(X, Y, W, Z) and it is symmetric between the first pair and last pair

Rm(X,Y,Z,W ) = Rm(Z, W, X, Y ).

Hence the coefficient of t2 in f(t) is

2Rm(X,Y,Z,W ) − 2Rm(Z,X,Y,W ). (1)

Interchanging X and Y , we similarly obtain that the expression

2Rm(Y,X,Z,W ) − 2Rm(Z,Y,X,W ), is determined by the sectional curvatures of M at p. Again using symmetry of Rm, this is

− 2Rm(X,Y,Z,W ) + 2Rm(Y,Z,X,W ). (2)

Recall also the algebraic Bianchi identity: X Rm(X,Y,Z,W ) = 0. all cyclic permutations of (X,Y,Z)

Using this, (1) − (2) = 6Rm(X,Y,Z,W ), so the Riemann curvature tensor is determined by the sectional curvature.

Definition 15.5. A M has if its sectional curvature function is constant, that is there exists a constant k such that secp(π) = k for all p ∈ M and all tangent planes π ⊂ TpM.

Remark 15.6. Let p ∈ M. If the sectional curvature function at p is zero (that is, secp = 0), then the curvature Rm = 0 at p.

Hence, a Riemannian manifold (M, g) is flat if and only if the sectional curvature is identically zero.

Let us consider the special case when our Riemannian manifold is a surface.

In that case we had already an intrinsic notion of curvature, namely the Gauss curvature.

3 Proposition 15.7. For a Riemannian surface (M, g), the sectional curvature of M is equal to its Gauss curvature.

Proof: By definition, for p ∈ M 2,

Rm2112 sec(TpM) = 2 (g11g22 − g12) h(∇ (∇ ∂ ) − ∇ (∇ ∂ )), ∂ i = ∂2 ∂1 1 ∂1 ∂2 1 2 , det g which is a formula for the Gauss curvature.

Exercise 15.8. Prove this expression, using

• the definition of Gauss curvature K(p) as the determinant of the shape operator −dNp, where N is the Gauss

det(II) • the resulting expression K = det I where I and II denote the 1st and 2nd fundamental forms. (You should remind yourself how to obtain this expression.)

• the fact (which you should make sure you understand) that

k ∇∂i (∂j) = Γij∂k + hijN

where hij denotes the coefficients of the 2nd fundamental form.

We will next study some which have constant sectional curvature, for which the follow- ing remark will prove helpful.

Remark 15.9. Let (M, g) be a Riemannian manifold and consider the (0, 4)-tensor Q defined by Q(X,Y,Z,W ) = hX,W i hY,Zi − hY,W i hX,Zi for X, Y, W, Z ∈ X (M). Then M has constant sectional curvature equal to K0 if and only if

Rm = K0Q, where Rm denotes the Riemann curvature tensor.

Proof:

Suppose M has constant sectional curvature K0 and take p ∈ M and linearly independent

X,Y ∈ TpM. Then

2 2 2 Rm(X,Y,Y,X) = K0(|X| |Y | − hX,Y i )

= K0Q(X,Y,Y,X).

4 As in the above proof, Lemma 15.4, this then implies that

Rm(X,Y,Z,W ) = K0Q(X,Y,Z,W ) for all X,Y,Z,W ∈ X (M), as required.

The converse is trivial.

Corollary 15.10 (Constant Sectional Curvature). Let (M, g) be a Riemannian manifold, p ∈ M and {e1, . . . , en} an orthonormal basis of TpM. Then the following are equivalent:

1. sec(π) = K0 for all 2-planes π ⊂ TpM

2. Rmijkl = K0(δilδjk − δikδjl).

3. Rmijji = −Rmijij = K0 for all i 6= j and Rmijkl = 0 otherwise.

5